Problem 59 - Single Variable Calculus

Question

find the given integral. \(\int \tanh x d x\)

Step-by-Step Solution

Answer

The short answer to the given integral is \(\int \tanh x \,d x = \ln|\cosh x| + C\).

Step 1: Rewrite the expression in terms of sinh and cosh functions

Rewrite the given integral in terms of hyperbolic sine and cosine functions: \[ \int \tanh x \,d x = \int\frac{\sinh x}{\cosh x} \,d x . \]

Step 2: Use substitution

Let's perform a substitution. Let \(u = \cosh x\). Then, \(du = \sinh x \,d x\). With this substitution, the integral becomes: \[ \int\frac{\sinh x}{\cosh x} \,d x = \int\frac{1}{u} \,d u. \]

Step 3: Integrate the new expression

Now, integrate the simplified expression with respect to \(u\): \[ \int\frac{1}{u} \,d u = \ln|u| + C. \]

Step 4: Replace u by the original function

Now that we have the antiderivative in terms of \(u\), we substitute back the original function to get our answer in terms of \(x\). We know that \(u = \cosh x\), so the final result is: \[ \ln|\cosh x| + C. \] Therefore, the antiderivative of the given integral is \(\ln|\cosh x| + C\).

Key Concepts

Understanding Hyperbolic Functions

Hyperbolic functions are similar to trigonometric functions, but are based on hyperbolas instead of circles. They are essential in various fields including calculus, engineering, and physics. The key hyperbolic functions include:

• Sinh (sine hyperbolic): \( \sinh x = \frac{e^x - e^{-x}}{2} \)


• Cosh (cosine hyperbolic): \( \cosh x = \frac{e^x + e^{-x}}{2} \)


• Tanh (tangent hyperbolic): \( \tanh x = \frac{\sinh x}{\cosh x} \)

These operations define shapes like the hyperbolic paraboloid often seen in architectural structures. Knowing these functions allows us to express complex equations in forms that are easier to integrate, differentiate, or otherwise manipulate.

Using the Substitution Method

The substitution method is a powerful tool in calculus that simplifies complex integrals. This technique involves substituting a part of the integral with a new variable, making it easier to evaluate. Here's how it works:

• Identify a section of the integral that can be replaced with a single variable \( u \).


• Differentiate this section to find \( du \), which represents the differential change.


• Substitute both the section and its differential into the integral, simplifying it significantly.

For our example, substituting \( u = \cosh x \) simplifies the integral of \( \tanh x \) into an easier form. Always remember to substitute back the original expression at the end, to reflect the variable in your solution.

Finding Antiderivatives

Antiderivatives, or indefinite integrals, are operations that reverse differentiation. They help us find functions whose derivative matches a given function. To compute an antiderivative:

• Identify the form of the integrand and consider any possible substitutions.


• Use known antiderivatives and properties, such as \( \int \frac{1}{u} \, du = \ln|u| + C \), as building blocks.


• Perform the integration, apply constants of integration like \( C \), which accounts for any constant added when differentiating.

In our example, integrating \( \frac{1}{u} \) results in \( \ln|u| + C \), and substituting \( u = \cosh x \) gives the final solution: \( \ln|\cosh x| + C \). Thus, antiderivatives convert expressions into functions, illustrating the inverse nature of integration to differentiation.